Momentum problem with collision

In summary, the queue ball's final speed is 2.1 m/s at an angle of theta with respect to its original line of motion.
  • #1

lzh

111
0

Homework Statement


Assume an elastic collision (ignoring friction
and rotational motion).
A queue ball initially moving at 3.7 m/s
strikes a stationary eight ball of the same size
and mass. After the collision, the queue ball's
final speed is 2.1 m/s at an angle of theta with
respect to its original line of motion.

http://img506.imageshack.us/img506/3229/untitledjb3.png [Broken]
Find the eight ball's speed after the colli-
sion. Answer in units of m=s.

Homework Equations


p=mv
possibly energy equations


The Attempt at a Solution


This problem would've been a lot easier if the problem provided theta's value. Without theta i only managed to get to this:
3.7=2.1cos(theta)+Vx
where Vx is the x component of the 8 ball after collision
0=2.1sin(theta)+Vy
where Vy is the y component of the 8 ball after collision
should i find theta first? or can I find one of the component's value first and then find theta?
 
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  • #2
lzh said:
0=2.1sin(theta)+Vy

First of all, you missed a sign.
 
  • #3
that doesn't matter does it?
just because Vy is negative you don't need to do this:
0=2.1sin(theta)-Vy
since that would end up having Vy's value as positive
 
  • #4
Vy is positive, and that is why you have to put the negative sign in front.

Although you haven't explicitly written it down, it's the components of the velocity vectors that you are equating.

So, you're second equation is something like
0j=2.1sin(theta)j-Vyj
 
  • #5
j is the mass of the ball?
if so I left them out because they cancel
 
  • #6
lzh said:
j is the mass of the ball?
if so I left them out because they cancel

By j, I meant a unit-vector in the direction parallel to the y-axis.

0j=2.1sin(theta)j-Vyj
=> 0j=(2.1sin(theta)-Vy)j

Since the vectors are equal, so are their magnitudes, and that's how you get
0=2.1sin(theta)-Vy
 
  • #7
oh, but I don't really see how that helps me in the problem
 
  • #8
hey guys may b u shud consider forming equs fom the line of collision...tats how they did in our class...im not v comfertable wit it either, but ill get ma note book if u want
 
  • #9
you mean breaking it into components? and having equation for x direction and y direction?
 
  • #10
well, in a similar prob, v got the first equs by e=v(seperation)/v(approach)...the other relations were used on the basis of LOC
 
  • #11
yea ...sumthing like tat...shud i get ma notebook?
 
  • #12
this problem supposedly should use momentum formulae. And i don't think I've ever seen your way of doing it? I sort of don't get it... what exactly does e stand for here and what is LOC.
yeah please get your notebook
 
  • #13
e=co-eff of restitution, n LOC is line of collision, , it says here Newtons law of collision can only b applied along the loc
 
  • #14
ummm...u shud know tat I am living in bombay, so even if the laws of physics don't change, the methods of attacking a ques do change ...n I am currently trainig for iit entrance exam...so, idk wht level u r at
 
  • #15
ah now i sort of get it, though they never taught about coefficient of restitution in school yet. So basically once i find the coeff of restitution all i need to do is subtract it by 2.1 m/s.
 
  • #16
now, for any spherrical bodies, loc is the line joining the centres of the two coliding bodies...u take the components of the resulting velocities along the loc, get the equs n solve...now i can giv u the solution inna while, but my personal advise wud b ,dont listen to me atall if the concept of loc is entirely new to u...but frankly, i don't c how the prob can b tackled otherwise
 
  • #17
not quite, u on msn?
 
  • #18
but there are 2 unknowns for this case, the velocity of 8 ball after collision and the coeff of restitution
 
  • #19
dear, u just said tat its a elastic collision, n to avoid friction n all, u ment a perfectly elastic collision, in this case, e=1, always
 
  • #20
whcih brings us to the eqn u already hav, u1= v1cos(theta)- v2...the udder eqs, i recommend u get aquainted wit the concept of loc
 
  • #21
oh sorry i have not been looking on the second page, so my previous few posts might not have made sense
 
  • #22
wht class r u in?10th or 11th?
 
  • #23
i'm using the forumula here:
http://www.racquetresearch.com/coeffici.htm [Broken]
so I basically set the equation equal to 1 and solve for the unknown velocity?
 
Last edited by a moderator:
  • #24
jay ambekar said:
wht class r u in?10th or 11th?
I'm in 10th
 
  • #25
bcoz I am in 11th , n they did e n loc whn they did collisions
 
  • #26
wow, ok...so according to me , u need to b familiar wit loc n e n all, but if they r giving a prob at ur level, witout the knowledge of dis, they probably expect u to manage witout loc...so basiclly by blabaring more, ill only confuse u more n nothing else...but if u want, ill ask ma proff tomorrow in class, since v r doing advance Newtonian phy...ill ask him...
 
  • #27
so i hope u can wait till tomorrow...ill get u the answer...giime ur is where i can mail, or udder way to contact...ill do my best...i got to leave for a snooze now
 
  • #28
yea, the link is gud, but the whole bounce height thing is not required rite now...bye for now, coz its 2am in India n i got to day ahead
 
  • #29
oh well this is due in like 9 hrs, so hopefully that's plenty of time.
and email is
jackz15@gmail.com
and about that link, besides having a forumula for bouncing off floor it also have one for collision
 
  • #30
ouch , i cud get this in like another 20 hrs, I am sorry, ma heads got too heavy wit sleep to think, ...im j.purpleink@gmail.com...well, don't worry too much...gud nite
 
  • #31
I would say that this collision is to be modeled as a perfectly elastic collision, and so you can use conservation of kinetic energy as well as conservation of momentum.

(This also means you don't need to worry about the coefficient of restitution, which I gather you haven't been taught).
 
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  • #32
if i use conservation of energy why do i need to use conservation of momentum? I know that i should use both but:
before collision: all kinetic energy
after:all kinetic
theres no friction and everything is perfect so if i set the kinetic energy from before equal to after I would just have one unknown(the answer) to solve.
1/2mv^2=1/2mv^2+1/2mv^2
basically the 1/2 and mass all cancel
so:
3.7^2=2.1^2+V^2
V=3.04
this definitely doesn't seem right...
 
  • #33
yeah i tried what was above and it was right... w/o any momentum...strange
 
  • #34
hey, check ur gmail inbox
 

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